please solve all parts (a-f) Consider the vectors:12and t222(a) Find the angle between these two vectors.(b) Find the vector projection 7, of v onto t.(c) Verify that = 7 - P is perpendicular to t.(d) Using the same logic, find a vector perpendicular to v.(e) Verify that Pythagoras law holds for t and t.(f) Verify the Schwarz inequality for d and .

Answered: Consider the vectors: 1 and 7 %3D 2 2…

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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please solve all parts (a-f)

Consider the vectors:
1
2
and t
2
2
2
(a) Find the angle between these two vectors.
(b) Find the vector projection 7, of v onto t.
(c) Verify that = 7 - P is perpendicular to t.
(d) Using the same logic, find a vector perpendicular to v.
(e) Verify that Pythagoras law holds for t and t.
(f) Verify the Schwarz inequality for d and .

Expert Solution

Step 1

Since we only answer up to 3 sub-parts, we’ll answer the first 3. Please resubmit the question and specify the other subparts (up to 3) you’d like answered

Step 2

Given: The vectors $\vec{v} = [\begin{matrix} 1 \\ 2 \\ 0 \\ 2 \end{matrix}]$ and $\vec{u} = [\begin{matrix} - 1 \\ 0 \\ 2 \\ 2 \end{matrix}]$ ,

To find:

(a) Angle between given vectors,

(b) vector projection $\vec{p}$ of $\vec{v}$ onto $\vec{u}$ .

To verify:

(c) $\vec{n} = \vec{v} - \vec{p}$ is perpendicular to $\vec{u}$ .

Step 3

Explanation:

(a)

It is known that the angle between two vectors $\vec{u} and \vec{v}$ is equal to $θ = \cos^{- 1} \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|}$ ,

Therefore angle between given vectors is

$θ = \cos^{- 1} (\frac{- 1 + 0 + 0 + 4}{\sqrt{1^{2} + 2^{2} + 0^{2} + 2^{2}} \sqrt{{(- 1)}^{2} + 0^{2} + 2^{2} + 2^{2}}}) \Rightarrow θ = \cos^{- 1} (\frac{3}{\sqrt{9} \sqrt{9}}) \Rightarrow θ = \cos^{- 1} (\frac{1}{3})$

Answer: Angle between given vectors is $θ = \cos^{- 1} (\frac{1}{3})$ .

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